Question: Let $R$ be the region enclosed by the line $x=3$, the line $y=2$, and the curve $y=\dfrac{1}{x-1}$. $y$ $x$ ${y=\dfrac{1}{x-1}}$ $ R$ $ 2$ $\left(3,\dfrac 1 2\right)$ $3$ A solid is generated by rotating $R$ about the line $x=3$. Which one of the definite integrals gives the volume of the solid? Choose 1 answer: Choose 1 answer: (Choice A) A $\pi \int_{\frac 12}^2 \left(3-\dfrac 1y\right)^2dy$ (Choice B) B $\pi \int_{\frac 12}^2 \left(3-\dfrac{1}{y-1}\right)^2dy$ (Choice C) C $\pi \int_{\frac 12}^2 \left(2-\dfrac{1}{y-1}\right)^2dy$ (Choice D) D $\pi \int_{\frac 12}^2 \left(2-\dfrac 1y\right)^2dy$
Solution: Let's imagine the solid is made out of many thin slices. $y$ $x$ ${y=\dfrac{1}{x-1}}$ Notice the slices are horizontal, because we are rotating $R$ about a vertical axis. Each slice is a cylinder. Let the thickness of each slice be $dy$ and let the radius of the base, as a function of $y$, be $r(y)$. Then, the volume of each slice is $\pi [r(y)]^2\,dy$, and we can sum the volumes of infinitely many such slices with an infinitely small thickness using a definite integral: $\int_a^b \pi [r(y)]^2\,dy$ This is called the disc method. What we now need is to figure out the expression of $r(y)$ and the interval of integration. Let's consider one such slice. $y$ $x$ ${y=\dfrac{1}{x-1}}$ $ 2$ $\left(3,\dfrac 1 2\right)$ $3$ $r$ The radius is equal to the distance between the curve ${y=\dfrac{1}{x-1}}$ and the line ${x=3}$. To find it, we need to solve $y=\dfrac{1}{x-1}$ for $x$ : ${x=\dfrac 1y+1}$ So, for any $y$ -value, this is the equation for $r(y)$ : $\begin{aligned} r(y)}&={3}-\left({\dfrac 1y+1}\right) \\\\ &=2-\dfrac 1y} \end{aligned}$ Now we can find an expression for the area of the cylinder's base: $\begin{aligned} &\phantom{=}\pi [r(y)}]^2 \\\\ &=\pi\left(2-\dfrac 1y}\right)^2 \end{aligned}$ The bottom endpoint of $R$ is at $y=\dfrac 12$ and the top endpoint is at $y=2$. So the interval of integration is $\left[\dfrac 12,2\right]$. Now we can express the definite integral in its entirety! $\begin{aligned} &\phantom{=}\int_{\frac 12}^2 \pi\left(2-\dfrac 1y\right)^2dy \\\\ &=\pi \int_{\frac 12}^2 \left(2-\dfrac 1y\right)^2dy \end{aligned}$